4
$\begingroup$

enter image description here

So I've been back and forth on whether linearity and constant variance assumptions are satisfied by these plots (for a linear model). What's throwing me off is how in the first plot above we have like a gaussian circle and also the region of highest point density seems to shift upwards (which affects the mean trend line a bit in red).

My professor tells me that these plots do not violate the assumptions as there is no significant change in the y-axis means over the x-axis. Is it really that simple in looking at these kinds of plots? The book we are using says that for linearity we want to see the points bounce randomly around the y=0 in a symmetric way in the first plot. For constant variance, we want to see a vertically uniform distribution of residuals in that plot. And for Scale-Location we just want to see that there is no major upward or downward trend.

Still, something about these plots bother me. I don't quite see linearity and something seems off about saying we have constant variance (move away to the left or right from the center of the cluster and it looks to me that variance starts to compress).

Can someone give me a bit more input and explanation on why linearity and constant variance assumptions hold in this case?

Update

Here is the actual data with the SLR line fitted and the lowess line superimposed:

enter image description here

And adding $\beta_2 x^2$ as suggested:

enter image description here

And $R^2$ goes up from 2.38% to 3.03%.

Improve this question
$\endgroup$

1 Answer 1

Reset to default
3
$\begingroup$

I am not entirely in agreement about the first plot. Usually, if your lowess fit line shows a pretty clear non-linear trend like this, there could be non-linear relationship between y and any of the x's unaccounted for. And that bending happens at a high data density area makes me feel even stronger about this. You can try compose some "component plus residual plot (URL)" to see which one is the culprit. And try adding a quadratic term of that variable into the model and see if it improves the overall fit.

I do, however, agree that the variance is reasonable. You're correct that in the second figure the dots seem to narrowing down from left to right. But also remember that most variables tend to have fewer extreme values than those closer to its central tendency (think the tails of a normal curve). So, as fitted value goes up, you'd expect fewer cases exist. In real world data a perfectly "block-like" appearance of variance plot seldom exists, unless the outcome is from some distribution like a uniform distribution.

Improve this answer
$\endgroup$
6
  • $\begingroup$ I should have added this is for a simple linear regression (there is only one x) $\endgroup$
    – Matt
    Commented Feb 27, 2018 at 15:43
  • $\begingroup$ In that case a scatter plot with the dependent at y axis, and the predictor at x axis, overlay with both linear fit line and lowess line should get you a pretty good clue. If adding a $x^2$ does improve the model, then it's no longer simple linear regression. If your assignment does not allow going outside the bound, then you can make a judgment call. $\endgroup$
    – Penguin_Knight
    Commented Feb 27, 2018 at 15:52
  • $\begingroup$ Yeah this is for SLR only (MLR will come in the next assignment). I've added the scatterplot with SLR line and lowess to the op. $\endgroup$
    – Matt
    Commented Feb 27, 2018 at 16:33
  • $\begingroup$ Seeing the new plot... I'd at least try to fit $y = \beta_0 + \beta_1 x + \beta_2 x^2$ and check what if $\beta_2$ is a significant predictor. $\endgroup$
    – Penguin_Knight
    Commented Feb 27, 2018 at 20:11
  • $\begingroup$ $\beta_1=0.66, p < 0.001$ and $\beta_2=-0.002, p < 0.01$. So yes significant but small effect size. $\endgroup$
    – Matt
    Commented Feb 28, 2018 at 3:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.